All categories

1 year ago

In △XYZ, m∠Z = 34, x = 61 cm, and z = 42 cm. Find m∠X. Round your answer to the nearest tenth of a degree.

1 answer:

8 0

M∠X = 54.3°.

Using the Law of Sines, we have:
$\frac{\sin{Z}}{z}=\frac{\sin{X}}{x}
\\
\\\frac{\sin{34}}{42}=\frac{\sin{X}}{61}$

Cross multiplying gives us
61(sin 34) = 42(sin X)

Divide both sides by 42:
(61(sin 34))/42 = (42(sin X))/42
(61(sin 34))/42 = sin X

Take the inverse sine of both sides:
sin⁻¹((61(sin 34))/42) = sin⁻¹(sin X)
54.3 = X

You might be interested in

Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s).

Arada [10]

If $y=3x^2+30x+71$ , then
$y=3(x^2+10x)+71=3(x^2+2\cdot5x+25-25)+71$
$y=3((x+5)^2-25)+71=3(x+5)^2-75+71=3(x+5)^2-4$ . So, in the first gap you can write 5 and in the second you can write -4.
Since $(x+5)^2\ge 0$ , the minimum value of y is when $(x+5)^2=0$ . The solution of the last equation is x=-5 and then y=-4 -- these are numbers, which you can write in the third and fourth gaps.

5 0

1 year ago

Read 2 more answers

A ball is thrown down from a balcony with a vertical velocity (in ft/s) given by

fenix001 [56]

Answer:

–90 < –32t – 10 < –58

Step-by-step explanation:

We want the velocity to be BETWEEN -90 and -58.

Whenever we need a quantity, let it be A, to be between two numbers, p and q (p is less than q), we can write it as:

p < A < q

Similarly, here we need the velocity, -32t-10 to be BETWEEN -90 and - 58 (with -90 being the smaller number). Thus we can write:

–90 < –32t – 10 < –58

This is the correct choice, 2nd choice.

4 0

2 years ago

Read 2 more answers

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3

In-s [12.5K]

Answer:

a) There is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

c) There is a 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3 minutes. This means that $\mu = 8.3, \sigma = 3.3$ .

(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?

We are working with a sample mean of 37 jets. So we have that:

$s = \frac{3.3}{\sqrt{37}} = 0.5425$

Total time of 320 minutes for 37 jets, so

$X = \frac{320}{37} = 8.65$

This probability is the pvalue of Z when $X = 8.65$ . So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{8.65 - 8.3}{0.5425}$

$Z = 0.65$

$Z = 0.65$ has a pvalue of 0.7422. This means that there is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?

Total time of 275 minutes for 37 jets, so

$X = \frac{275}{37} = 7.43$

This probability is subtracted by the pvalue of Z when $X = 7.43$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{7.43 - 8.3}{0.5425}$

$Z = -1.60$

$Z = -1.60$ has a pvalue of 0.0548.

There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?

Total time of 320 minutes for 37 jets, so

$X = \frac{320}{37} = 8.65$

Total time of 275 minutes for 37 jets, so

$X = \frac{275}{37} = 7.43$

This probability is the pvalue of Z when $X = 8.65$ subtracted by the pvalue of Z when $X = 7.43$ .

So:

From a), we have that for $X = 8.65$ , we have $Z = 0.65$ , that has a pvalue of 0.7422.

From b), we have that for $X = 7.43$ , we have $Z = -1.60$ , that has a pvalue of 0.0548.

So there is a 0.7422 - 0.0548 = 0.6874 = 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.

7 0

2 years ago

Tomas is a commercial real estate agent who earns a yearly salary of $82,500 and a 2.5% commission on his total sales. If Tomas

Georgia [21]

Tomas earning is $176,250

7 0

1 year ago

Read 2 more answers

Problem Lois and Clark own a company that sells wagons. The amount they pay each of their sales employees (in dollars) is given

Oxana [17]

12h + 30w.....where h = hrs worked and w = wagons sold

so if an employee works 6 hrs and sells 3 wagons....then h = 6 and w = 3

12h + 30w
12(6) + 30(3) =
72 + 90 = $ 162 <==

8 0

2 years ago